Assume that y is a function of x, find dy/dx given that tan^-1 (x y) = x + y^3

Question

Derivative of inverse trigonometric functions.

Answer 1

tan⁻¹(xy)=x+y³,

In the derivative y'=dy/dx:

(xy'+y)/(1+x²y²)=1+3y²y',

xy'+y=(1+x²y²)(1+3y²y')=1+3y²y'+x²y²+3x²y⁴y',

xy'-3y²y'-3x²y⁴y'=1+x²y²-y,

(x-3y²-3x²y⁴)y'=1+x²y²-y,

dy/dx=(x²y²-y+1)/(x-3y²-3x²y⁴).